Question

The points A(4,7) B(p,3) and C(7,3) are the vertices of a right triangle, right -angled at B, Find of P.

Answer 1

REF.Image

A(4,7), B(p,3) & c (7,3) are vertices of right angle triangle then u

using Pythagoras theorem

$(AC{)}^2=(AB{)}^2+(BC{)}^2$

$(7-4{)}^2+(3-7{)}^2=(p-4{)}^2+(3-7{)}^2+(p-7{)}^2+(3-3{)}^2$

$9+16=(p-4{)}^2+16+(p-7{)}^2$

$9=p^2+16-8p+p^2+49-14p$

$2p^2-22p+56=0$

$p^2-11p+28=0$

$p^2-7p-4p+28=0$

(p-4)(p-7)=0

Hence p-4=0 & p-7=0

[p=4,7]